Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> tree[1 << 20], cities[1 << 20], st[1 << 22];
set<int> ans;
int chainNo, chainInd[1 << 20], chainHead[1 << 20], ptr, basearray[1 << 20];
int par[20][1 << 20], depth[1 << 20], subsize[1 << 20], posInBase[1 << 20];
void dfs(int cur, int prev, int _depth = 0) {
  par[0][cur] = prev;
  depth[cur] = _depth;
  subsize[cur] = 1;
  for (int i = 0; i < tree[cur].size(); i++) {
    int to = tree[cur][i];
    if (to != prev) {
      dfs(to, cur, _depth + 1);
      subsize[cur] += subsize[to];
    }
  }
}
void HLD(int cur, int prev) {
  if (chainHead[chainNo] == -1) chainHead[chainNo] = cur;
  chainInd[cur] = chainNo;
  basearray[ptr] = cur;
  posInBase[cur] = ptr++;
  int sc = -1;
  for (int i = 0; i < tree[cur].size(); i++)
    if (tree[cur][i] != prev)
      if (sc == -1 || subsize[sc] < subsize[tree[cur][i]]) sc = tree[cur][i];
  if (sc != -1) HLD(sc, cur);
  for (int i = 0; i < tree[cur].size(); i++)
    if (tree[cur][i] != prev)
      if (sc != tree[cur][i]) {
        chainNo++;
        HLD(tree[cur][i], cur);
      }
}
int lca(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  int diff = depth[u] - depth[v];
  for (int i = 0; i < 20; i++)
    if ((diff >> i) & 1) u = par[i][u];
  if (u == v) return u;
  for (int i = 20 - 1; i >= 0; i--)
    if (par[i][u] != par[i][v]) {
      u = par[i][u];
      v = par[i][v];
    }
  return par[0][u];
}
void build(int node, int start, int end) {
  if (start == end) {
    for (int i = 0; i < cities[basearray[start]].size(); i++)
      st[node].push_back(cities[basearray[start]][i]);
    sort(st[node].begin(), st[node].end());
    if (st[node].size() > 10) st[node].resize(10);
    return;
  }
  int left = node << 1;
  int right = left | 1;
  int mid = (start + end) >> 1;
  build(left, start, mid);
  build(right, mid + 1, end);
  for (int i = 0; i < st[left].size(); i++) st[node].push_back(st[left][i]);
  for (int i = 0; i < st[right].size(); i++) st[node].push_back(st[right][i]);
  sort(st[node].begin(), st[node].end());
  if (st[node].size() > 10) st[node].resize(10);
}
void query(int node, int start, int end, int left, int right) {
  if (left >= start && right <= end) {
    for (int i = 0; i < st[node].size(); i++) ans.insert(st[node][i]);
    return;
  }
  if (left > end || right < start) return;
  int mid = (left + right) >> 1;
  int lnode = node << 1;
  int rnode = lnode | 1;
  query(lnode, start, end, left, mid);
  query(rnode, start, end, mid + 1, right);
}
void solve(int u, int v) {
  while (chainInd[u] != chainInd[v]) {
    query(1, posInBase[chainHead[chainInd[u]]], posInBase[u], 0, ptr - 1);
    u = par[0][chainHead[chainInd[u]]];
  }
  query(1, posInBase[v], posInBase[u], 0, ptr - 1);
  return;
}
int main() {
  int n, m, q;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &q);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d", &u);
    scanf("%d", &v);
    u--;
    v--;
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    int temp;
    scanf("%d", &temp);
    temp--;
    cities[temp].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    chainHead[i] = -1;
    for (int j = 0; j < 20; j++) par[j][i] = -1;
  }
  chainNo = 0;
  dfs(0, -1);
  HLD(0, -1);
  build(1, 0, ptr - 1);
  for (int i = 1; i < 20; i++)
    for (int j = 0; j < n; j++)
      if (par[i - 1][j] != -1) par[i][j] = par[i - 1][par[i - 1][j]];
  par[0][0] = -1;
  while (q--) {
    int u, v, a, l;
    scanf("%d", &u);
    scanf("%d", &v);
    scanf("%d", &a);
    u--;
    v--;
    l = lca(u, v);
    solve(u, l);
    solve(v, l);
    if (ans.size() < a) {
      printf("%d ", ans.size());
      for (auto it = ans.begin(); it != ans.end(); it++) printf("%d ", *it + 1);
      puts("");
    } else {
      printf("%d ", a);
      set<int>::iterator it = ans.begin();
      while (a--) {
        printf("%d ", *it + 1);
        it++;
      }
      puts("");
    }
    ans.clear();
  }
  return 0;
}